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ROTATIONAL MOTION Mrs. CHITRA JOSHI PGT (Physics) CENTRE OF MASS AND ROTATIONAL MECHANICS System: A collection of any number of particles interacting with one another are said to form a system. Concept of centre of mass: Centre of mass of a body or a system of bodies as the point that moves as though all of the were concentrated there and all external forces were applied there. Centre of mass of a two particle Let m1, m2 are masses R1,R2 position vector of mass m. R of centre of mass is given by Centre of mass of two particle system always lies between the particle (m1&m2) coordinate of m1 are x1y coordinate of m2 are x2y2 coordinate of C.M. x = (m1x1 + m2x2)/m1+m2 y = (m1y1 + m2y2)/m1 + m2 m1= m2= m If m1 = m2 Then x = m(x1 + x2)/2m = (x1 + x2)/2 C.M lies at the mid point of the line joining m1,m2 (when m1= m2= m) Momentum conservation of centre of mass equation of motion of centre of mass given by Fext = dmv/dt fext = 0 d(mv) = 0 dt mv = constant If no external force is act on system momentum is constant is called law of conservation of momentum. Eg. A special type of cracker which when ignited goes up in the sky and then expands in the mid air into many firely pieces. As the cracker expands under the action of internal forces the centre of mass of 4 fragment should continue moving along the same parabolic path. The centre of mass of 4 fragments will move on initial parabolic path PBC, which is the continuation of the initial parabolic path AP of the cracker. Rigid body : A body is taken as rigid, where change in inter particle distances under the effect of external forces can be ignored. Center of mass of a rigid body : The center of mass of a rigid body is defined as a point where the entire mass of the rigid body is supposed to be concentrated. The nature of motion of the rigid body shall remain unaffected, if all the forces acting on the body were applied directly on the center of mass of the body. s. no. Body Position of center of mass 1. 2. 3. 4. 5. 6. Uniform hollow sphere Uniform solid sphere Uniform circular ring Uniform circular disc Uniform rod A plane lamina in the form of a square or a rectangle or a parallelogram. Triangular plane lamina Center of the sphere Center of the sphere Center of the ring Center of the disc Center of the rod Points of intersection of diagonals. Points of intersection of the medians of the triangle. Points of intersection of the diagonals. Middle point of the axis of the cylinder. Middle point of the axis of the cylinder. On the axis of the cone at a point distant 3h/4 from the vertex O where h=OA is height of the cone. 7. 8. 9. Rectangular or cubical block. Hollow cylinder 10. Solid cylinder 11. Cone or pyramid Rotational motion of a particle in a plane and concept of Torque Torque:- It is a product of force and perpendicular distance. τ = F*d Unit of torque is Nm [τ ] = [ML2T-2] Expression of torque in Cartesian coordinate system W = F.dr or W = τ . dθ = (Fxî +Fyĵ) * (dxî + dyĵ) W = Fxdx + Fydy --------------(1) [(d/dθ)*cosθ = - sinθ] [(d sinθ/dθ) = cosθ] x = r cosθ dx/dθ = -r sinθ dx = -ydθ ------------(2) (because y = r sinθ) dy/dθ = r cosθ dy = xdθ -------------(3) From (1),(2) &(3) W = F*dy + Fydy = F*(-ydθ) + Fy(xdθ) = (xFy – yFx)dθ -------(4) dW = τdθ -----------------(5) From (4)&(5) τdθ = (xfy – yfx)dθ dω = τdθ (rotational motion) dω/dt = τdθ/dt (linear motion) P = τω If we apply force in x,y plane torque will be along z axis. Expression of Torque in polar coordinate system Fx = F cos α Fy = F sin α α=θ+ ø Ø=α-θ x = r cosθ y = r sinθ τ = xfy – yfx = r cosθ. Fsinα – r sinθ.F cosα = rf (sinα.cosθ – cosα. sinθ) = rf sin (α – θ) = rf sin ø = f(r sin ø) = F*d Angular momentum: If we can represent torque as the rate of change of some quantity, that quantity will be rotational analogue of linear momentum, and it can appropriately be called the angular momentum. L = r*p = r*mv Mkg*m/s Unit of angular momentum, L = Kgm2/s [L] = [ML2T-1] Expression for angular momentum In order to express torque as the rate of change of some quantity , we rewrite expression for torque rotating a particle in XY plane as τ = xFy-yFx If Px = mvx and Py = mvy are the x and y components of linear momentum of the body, then According to Newton’s 2nd law of motion, = Fy dPy/dt dPx/dt = Fx τz = d/dt*(xPy – yPx) ------------(1) τ = dl/dt -------------(2) [L + xPy – yPx] from (1)&(2) Px = p cosα py = p sinα α=θ+Ø Ø=α+θ x = r cos θ y = r sin θ From above Equation. L = xPy – yPx = r cos θ.P sin α – r sin θ.Pcos α = rP (cosθ.sinα – cosα. Sinθ) = rP sin (α - θ) L = rP sinØ Pθ = P sinØ-----------(a) Pr = p cosØ [L = Pθr] -----------(b) from (b) L = Pd Geometrical meanings of angular momentum Z M o X Y K L Area of parallelogram on OABC = r + dr Area of triangle OAB = ½*r*dr Velocity of particle = ν Time taken = dt V = dr/dt dr = v*dt dA = ½*r*v*dt We know that P = mv v = p/m dA = ½ r*P/m*dt dA/dt = L/2m [L =r*p] [dA/dt = Rate of change of are by time is called Areal velocity.] L = 2 mdA/dA Angular momentum is equal to the product of 2 mass and areal velocity. Rotational inertia or moment of inertia of the body: A quantity that measures the inertia of rotational motion of the body. Units of moment of Inertia As I = mass (distance)2 Therefore units of moment of inertia are Kg m2 or g cm. The dimensions of moment of inertia are [MLT]s Radius of Gyration: The radius of gyration of a body about a given axis is the perpendicular distance of a point P from the axis, where if whole mass of the body were concentrated, the body shall have the same moment of inertia as it has with the actual distribution of mass. This distance is represented by k. I = Mk2 ------------(1) = m1r2 + m2r22 …mnrn2 (m1 + m2 + m3 = M) I = m(r12 + r22 + r32 +…rn2) = {m*n (r12 + r22 + r32 +…rn2)/n} ------------(2) From (1)&(2) Mk2 = M (r12 + r22 + r32 +…rn2) K = [(r12 + r22 + r32 +…rn2)/n]1/2 ‘K’ is root mean square of distance of the particle from the axis of rotation Torque and moment of inertia consider a rigid body rotating bout a given axis with a uniform angular α, under the action of a torque. Let the body consist of particles of masses m1,m2,m3…mn at perpendicular distances r1,r2,r3…rn respectively from the axis of rotation As the the body is rigid , angular acceleration α of all the particles of the body is same . However there linear acceleration are different because of different distance of the particles form the axis. If a1, a2, a3…an are the respective linear acceleration of the particles, then a1 = r1α, a2 = r2 α, a3 =r3 α,… Force on the particle of mass m is F = m1α1 = m1r1α Moment of this force about the axis of rotation = f1 * r1 = (m1r1 α)*r1 = m1r12α Similarly, moments of force on other particles about the axis of rotation are m2r22 α, m3r32 α,…mnrn2 α. Therefore Torque acting on the body, τ = m1r2 α + m22r22 α + m3r32 α +…mnrn2α = (m1r12 + m2r22 + m3r32 +…mnrn2)α i=n τ = ( Σmir12 ) α i=1 or τ = I α i=n where I = ( Σ mir12 )moment of inertia of the body i=1 about the given axis of rotation. If α = 1, τ = I*1 or I = τ Hence moment of inertia of a body about a given the axis is numerically equal to torque acting on the body rotating with unit angular acceleration about it. We may rewrite equation (9) in vector form as τ =Iα This equation is called Fundamental equation of rotation or law of rotation.This corresponds to F = m α, which is the fundamental equation of linear motion. Principle of conservation of Angular momentum: According to this principle, when no external torque acts on a system of particles,then the total angular momentum of the system remains always a constant. Some examples of conservation of Angular momentum 1.An ice skater or a ballet dancer can increases her angular velocity by folding her arms and bringing the stretched legs close to other leg. 2.The driver jumping from the spring board sometimes exhibits somersaults in air before touching the water layer. Theorem of Perpendicular axis:- It state that the momentum of inertia of a plane lamina about an axis perpendicular to its plane is equal to the sum of moments of inertia of lamina about any two mutually perpendicular axis to its plane and intersecting each other at the point where the perpendicular axis passing through it. Theorem of parallel axis:- It state that the moment of inertia of a rigid body about any axis is equal to its moment of inertia about a parallel axis through its centre of mass and the product of mass of the body and the square of the perpendicular distance between the axis. Moment of inertia of a thin circular ring suppose M is the mass of a thin circular ring of radius R with centre O.we have to calculate moment of inertia of the ring about an axis YOY’,perpendicular to the plane of the ring and passing through its centre,Fig.5(b).12. Y O Y’ R dx Length of the ring = circumference = 2πR Mass per unit length of the ring = M/2πR Consider a small element of the ring of length dx Mass of this element = M dx 2πR Moment of inertia of this element about YOY’ = [(M/2πR)*dx] R2 =( MR/2π)*dx As the small element can lie anywhere over the entire length of the ring i.e., from x = 0 to x = 2πR, therefore, Moment of inertia of the entire circular ring about YOY’ I = MR2 = mass(radius)2 Moment of inertia of thin uniform Rod Suppose M is mass of a thin uniform rod of length l. We have to calculate moment of inertia of the rod about an axis YY` passing through the centre O of the rod and perpendicular to its length, Mass per unit length of the rod = M/l Consider a small element of length = dx, of the rod at a distance x from the centre. Mass of the element = M/l * dx Moment of inertia of this element about the given axis = mass x (distance)2 = M/l * dx (x 2) = M/l .x2 .dx As the element chosen can lie anywhere from left end of the rod (x = -l/2) to the right end of the rod (x = + /2), therefore, Moment of inertia of the uniform rod about YOY` is I = 1/12 Ml2 BODY 1. Uniform rod of length l. 2. Uniform rectangular lamina of length l and breadth b. 3. Uniform circular ring of radius of radius r. 4. Uniform circular disc of radius r. 5.Hollow cylinder of radius r. 6. Solid cylinder of radius r. 7.Hollow sphere of radius r. 8. Solid sphere of radius r. AXIS 1. Perpendicular to rod through its centre. 2.Perpendicular to lamina and through its centre. M.I. 1/12Ml M(l2 +b2)/12 3. Perpendicular to its MR2 plane through the centre 4.Perpendicular to its plane 2 and through the centre. ½ MR 5. Axis of cylinder 6. Axis of cylinder 7. Diameter MR2 ½ MR2 2/3 8. Diameter MR2 2/5 MR2 Linear Motion 1. Distance /displacement (s) Rotational Motion 1. Angle or angular displacement(θ) 2. Linear velocity, v = ds/ dt 2. Angular velocity ,ω = d θ/dt 3. Linear acceleration,a = dv/dt 3. Angular acceleration,a = dω/dt 4. Mass (m) 4. Moment of inertia (I) 5. Linear momentum p = mv 5. Angular momentum L = I ω 6. Force F = ma 6. Torque, τ = I ω 7. Also, force F = dp/dt 7. Also, torque τ = dL/dt 8. Translational K.E. = ½ mv2 8. Rotational K.E. = ½ I ω2 9. Work done, W = Fs 9. Work done, W = τθ 10. Power P = Fv 10. Power = τω 11. Linear momentum of a 11. Angular momentum of a system is system is conserved conserved when no external when no external force acts torque acts on the system. on the system. ( Principle of (Principle of conservation of conservation of linear momentum) angular momentum ) 12. Equation of translation motion (i) v = u + at (ii) S = ut +1/2 at2 (iii) v - u = 2as,where the symbols have their usual meaning. 13. Distance travelled in nth second ,snth = u+a/2(2n-1) 12. Equation of rotation motion (i) ω2 = ω1 + αt (ii) θ = ω1t + ½αt2 (iii) ω22 - ω12 = 2α θ, Where the symbols have their usual meaning 13 Angle traced in nth second. θnth = ω1 + α/2(2n –1)